BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A lecture course consists of 595 students. The students are

Expert replies
by BTGmoderatorLU » Thu May 09, 2019 11:50 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

Source: Veritas Prep

A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?

A. 17
B. 35
C. 45
D. 85
E. 119

The OA is C
Join the discussion
Source: — Problem Solving |

by Brent@GMATPrepNow » Thu May 09, 2019 12:05 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?

A. 17
B. 35
C. 45
D. 85
E. 119

The OA is C
In order to have an EQUAL number of students in each section, the number of students per section MUST BE A FACTOR of 595

Let's do some prime factorization
595 = (5)(7)(17)
From the prime factorization, we can see that answer choice A, B, D and E are all factors of 595

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by Scott@TargetTestPrep » Thu May 16, 2019 6:52 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?

A. 17
B. 35
C. 45
D. 85
E. 119

The OA is C
Any number whose digits sum to a number divisible by 3 is itself divisible by 3. For example, 3,912 is divisible by 3 because the sum of its digits is 3 + 9 + 1 + 2 = 15, which is divisible by 3.

Since 5 + 9 + 5 = 19, we see that 595 is not a multiple of 3. Since 45 is a multiple of 3, we cannot have 45 students in a discussion section.

Alternate solution:

Since 595 = 5 x 119 = 5 x 7 x 17, we see that 17 and 119 obviously can be the number of students in a discussion section, and so can 35 (which is 5 x 7) and 85 (which is 5 x 17). So, by process of elimination, it can't be 45.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Join the discussion